Lời giải:
Lấy PT(2) nhân 2 rồi cộng với PT(1) theo vế thu được:
$2(y-2x)+3x-2y=5.2+(-8)$

$\Leftrightarrow -x=2\Leftrightarrow x=-2$

$y=2x+5=2(-2)+5=1$

Vậy hpt có nghiệm $(x,y)=(-2,1)$

\(1,\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\2y+10+y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{16}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}3x=1-2y\\1-2y+y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\3y+6+2y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

e.

\(\left\{{}\begin{matrix}2x-3y+5=0\\3x+5y-21=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}10x-15y=-25\\9x+15y=63\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}19x=38\\3x+5y=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{21-3x}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)

f.

\(\left\{{}\begin{matrix}x-y\sqrt{2}=0\\2x\sqrt{2}+y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-y\sqrt{2}=0\\4x+y\sqrt{2}=5\sqrt{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x=5\sqrt{2}\\2x\sqrt{2}+y=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{2}\\y=5-2x\sqrt{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{2}\\y=1\end{matrix}\right.\)

a.

\(\Leftrightarrow\left\{{}\begin{matrix}5x=-25\\3x-5y=-30\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=\dfrac{3x+30}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=3\end{matrix}\right.\)

b.

\(\Leftrightarrow\left\{{}\begin{matrix}8x-6y=-10\\9x+6y=-24\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}17x=-34\\9x+6y=-24\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=\dfrac{-24-9x}{6}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=-1\end{matrix}\right.\)

c.

\(\left\{{}\begin{matrix}3x+3y=9\\4x-2y=-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y=3\\2x-y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x=2\\2x-y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=2x+1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=\dfrac{7}{3}\end{matrix}\right.\)

d.

\(\left\{{}\begin{matrix}5x-4y=32\\6x+2y=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x-4y=32\\12x+4y=36\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x-4y=32\\17x=68\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=\dfrac{3x-32}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-3\end{matrix}\right.\)

1. \(\left\{{}\begin{matrix}3x+4y=11\\2x-y=-11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+4y=11\\8x-4y=-44\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+4y=11\\11x=-33\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=5\\x=-3\end{matrix}\right.\)

2. \(\left\{{}\begin{matrix}3x+2y=0\\2x+y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+2y=0\\4x+2y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=-2\end{matrix}\right.\)

3.\(\left\{{}\begin{matrix}3x+\dfrac{5}{2}y=9\\2x+\dfrac{1}{3}y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+5y=18\\6x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4y=12\\6x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=3\\x=\dfrac{1}{2}\end{matrix}\right.\)

a, b và c có thể dùng phương pháp thế hoặc cộng trừ đại số

\(a,\left\{{}\begin{matrix}x=1-y\\1-y-y=-5\end{matrix}\right.=>\left\{{}\begin{matrix}x=1-y\\1-2y=-5\end{matrix}\right.=>\left\{{}\begin{matrix}x=1-y\\2y=6\end{matrix}\right.=>\left\{{}\begin{matrix}x=1-y\\y=3\end{matrix}\right.=>\left\{{}\begin{matrix}x=-2\\y=3\end{matrix}\right.\)

Kết luận hpt có 1 nghiệm duy nhất (x;y)=(-2;3)

b và c làm tương tự

a.\(\Leftrightarrow\left\{{}\begin{matrix}2x=-4\\x-y=-5\end{matrix}\right.\) ( cộng đại số bạn nhé )

\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\-2-y=-5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=3\end{matrix}\right.\)

b.\(\Leftrightarrow\left\{{}\begin{matrix}3x=6\\x-2y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\2-2y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)

c.\(\Leftrightarrow\left\{{}\begin{matrix}4x+6y=10\\9x-6y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}13x=13\\9x-6y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\9.1-6y=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

a, \(\left\{{}\begin{matrix}x+y=1\\x-y=-5\end{matrix}\right.\)

\(\Leftrightarrow x+y+x-y=-4\)

\(\Leftrightarrow2x=-4\)

\(\Leftrightarrow x=-2\)

Thay \(x=-2\) vào \(x+y=1\)\(\Leftrightarrow-2+y=1\)\(\Leftrightarrow y=3\)

Vậy \(x=-2;y=3\)

a: \(\left\{{}\begin{matrix}3x-2y=11\\4x-5y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3x=11+2y\\4x-5y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\4\left(\dfrac{2}{3}y+\dfrac{11}{3}\right)-5y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\\dfrac{8}{3}y+\dfrac{44}{3}-5y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}y+\dfrac{11}{3}\\-\dfrac{7}{3}y=3-\dfrac{44}{3}=-\dfrac{35}{3}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=5\\x=\dfrac{2}{3}\cdot5+\dfrac{11}{3}=\dfrac{10}{3}+\dfrac{11}{3}=\dfrac{21}{3}=7\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{y}{3}=1\\5x-8y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{y}{3}+1\\5x-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\5\left(\dfrac{2}{3}y+2\right)-8y=3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y+2\\\dfrac{10}{3}y+10-8y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{14}{3}y=3-10=-7\\x=\dfrac{2}{3}y+2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=7:\dfrac{14}{3}=7\cdot\dfrac{3}{14}=\dfrac{3}{2}\\x=\dfrac{2}{3}\cdot\dfrac{3}{2}+2=3\end{matrix}\right.\)

c: \(\left\{{}\begin{matrix}3x+5y=1\\2x-y=-8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x+8\\3x+5\left(2x+8\right)=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2x+8\\3x+10x+40=1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=2x+8\\13x=-39\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=-3\\y=2\cdot\left(-3\right)+8=8-6=2\end{matrix}\right.\)

d: \(\left\{{}\begin{matrix}\dfrac{x}{y}=\dfrac{2}{3}\\x+y-10=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}y\\x+y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{3}y+y=10\\x=\dfrac{2}{3}y\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{5}{3}y=10\\x=\dfrac{2}{3}y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=6\\x=\dfrac{2}{3}\cdot6=4\end{matrix}\right.\)

\(1,\Leftrightarrow\left\{{}\begin{matrix}x=2y+4\\-4y-8+5y=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\cdot5+4=14\\y=5\end{matrix}\right.\\ 2,\Leftrightarrow\left\{{}\begin{matrix}5x-30+6x=3\\y=10-2x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\\ 3,\Leftrightarrow\left\{{}\begin{matrix}x=4-2y\\6y-12+y=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{10}{7}\\y=\dfrac{19}{7}\end{matrix}\right.\)

a) Ta có: \(\left\{{}\begin{matrix}-x+2y=3\\3x+y=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-3x+6y=9\\3x+y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}7y=8\\-x+2y=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{8}{7}\\-x=3-2y=3-2\cdot\dfrac{8}{7}=\dfrac{5}{7}\end{matrix}\right.\)

hay \(\left\{{}\begin{matrix}x=-\dfrac{5}{7}\\y=\dfrac{8}{7}\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-\dfrac{5}{7}\\y=\dfrac{8}{7}\end{matrix}\right.\)

b) Ta có: \(\left\{{}\begin{matrix}2x+2\sqrt{3}\cdot y=1\\\sqrt{3}x+2y=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{3}x+6y=\sqrt{3}\\2\sqrt{3}x+4y=-10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2y=\sqrt{3}+10\\\sqrt{3}x+2y=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\sqrt{3}+10}{2}\\x\sqrt{3}+2\cdot\dfrac{\sqrt{3}+10}{2}=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{\sqrt{3}+10}{2}\\x\sqrt{3}=-5-\sqrt{3}-10=-15-\sqrt{3}\end{matrix}\right.\)

hay \(\left\{{}\begin{matrix}x=-1-5\sqrt{3}\\y=\dfrac{\sqrt{3}+10}{2}\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=-1-5\sqrt{3}\\y=\dfrac{\sqrt{3}+10}{2}\end{matrix}\right.\)

a, \(\left\{{}\begin{matrix}\\6x+2y=-2\end{matrix}\right.-6x+12y=18}\)

\(\left\{{}\begin{matrix}6\left(x+y\right)=8+2x-3y\\5\left(y-x\right)=5+3x+2y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x+6y=8+2x-3y\\5y-5x=5+3x+2y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}6x-2x+6y+3y=8\\-5x-3x+5y-2y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-8x+3y=5\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}4x+9y=8\\-24x+9y=15\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}28x=-7\\4x+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{7}{28}=-\dfrac{1}{4}\\4.\left(-\dfrac{1}{4}\right)+9y=8\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\\y=1\end{matrix}\right.\\ Vậy:\left(x;y\right)=\left(-\dfrac{1}{4};1\right)\)